NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.1#
Exercise 12.1 covers the surface area of combinations of solids — cubes, cuboids, cylinders, cones, hemispheres. The total surface area excludes the joined faces.
Key Formulas:
- Cylinder: Curved SA = $2\pi rh$, Total SA = $2\pi r(r+h)$
- Cone: Curved SA = $\pi rl$, Total SA = $\pi r(r+l)$, where $l = \sqrt{r^2+h^2}$
- Sphere: SA = $4\pi r^2$
- Hemisphere: Curved SA = $2\pi r^2$, Total SA = $3\pi r^2$
- Cube: SA = $6a^2$; Cuboid: SA = $2(lb+bh+hl)$
1. 2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid. #
Answer
Volume of each cube = 64 cm³ → side $a = 4$ cm.
When joined end to end: $l = 8$ cm, $b = 4$ cm, $h = 4$ cm.
$$\text{SA} = 2(lb + bh + hl) = 2(8 \times 4 + 4 \times 4 + 4 \times 8) = 2(32+16+32) = 2 \times 80 = 160 \text{ cm}^2$$2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel. #
Answer
Radius of hemisphere = radius of cylinder = 7 cm. Height of cylinder = $13 - 7 = 6$ cm.
Inner SA = Curved SA of cylinder + Curved SA of hemisphere
$$= 2\pi rh + 2\pi r^2 = 2\pi r(h+r) = 2 \times \frac{22}{7} \times 7 \times (6+7) = 44 \times 13 = 572 \text{ cm}^2$$3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy. #
Answer
Radius = 3.5 cm, height of cone = $15.5 - 3.5 = 12$ cm.
Slant height of cone: $l = \sqrt{r^2 + h^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5$ cm.
TSA = Curved SA of cone + Curved SA of hemisphere
$$= \pi rl + 2\pi r^2 = \pi r(l + 2r) = \frac{22}{7} \times 3.5 \times (12.5 + 7) = 11 \times 19.5 = 214.5 \text{ cm}^2$$4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid. #
Answer
Greatest diameter = 7 cm → radius = 3.5 cm.
SA of solid = SA of cube − base circle of hemisphere + Curved SA of hemisphere
$= 6 \times 7^2 - \pi r^2 + 2\pi r^2 = 294 + \pi r^2$
$= 294 + \dfrac{22}{7} \times 12.25 = 294 + 38.5 = \mathbf{332.5}$ cm²
5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid. #
Answer
Edge of cube = $l$, radius of hemisphere = $\dfrac{l}{2}$.
SA = SA of cube − circular base of hemisphere + curved SA of hemisphere
$$= 6l^2 - \pi\left(\frac{l}{2}\right)^2 + 2\pi\left(\frac{l}{2}\right)^2 = 6l^2 + \pi \cdot \frac{l^2}{4} = \frac{l^2}{4}(24 + \pi)$$6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. #
Answer
Radius = 2.5 mm, diameter = 5 mm.
Length of cylinder = total length − 2 × radius = $14 - 5 = 9$ mm.
SA = Curved SA of cylinder + 2 × Curved SA of hemisphere (one full sphere)
$$= 2\pi rh + 4\pi r^2 = 2\pi r(h + 2r) = 2\pi \times 2.5 \times (9 + 5) = 5\pi \times 14 = 70\pi = 70 \times \frac{22}{7} = 220 \text{ mm}^2$$7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹500 per m². (Note that the base of the tent will not be covered with canvas.) #
Answer
Radius = 2 m, cylinder height = 2.1 m, cone slant height = 2.8 m.
Canvas area = Curved SA of cylinder + Curved SA of cone
$$= 2\pi rh + \pi rl = \pi r(2h + l) = \frac{22}{7} \times 2 \times (2 \times 2.1 + 2.8) = \frac{44}{7} \times 7 = 44 \text{ m}^2$$Cost = $44 \times 500 =$ ₹22,000
8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm². #
Answer
Radius = 0.7 cm, h = 2.4 cm.
Slant height of cone: $l = \sqrt{r^2+h^2} = \sqrt{0.49+5.76} = \sqrt{6.25} = 2.5$ cm.
TSA = Curved SA of cylinder + Area of top circle + Curved SA of cone (inner hollowed) + Area of base
Actually: TSA = Curved SA of cylinder + Base circle + Curved SA of cone (hollow inside)
= $2\pi rh + \pi r^2 + \pi rl$
$= \pi r(2h + r + l)$
$= \dfrac{22}{7} \times 0.7 \times (4.8 + 0.7 + 2.5)$
$= 2.2 \times 8 = 17.6$ cm²
∴ TSA ≈ 18 cm² (to nearest cm²).
9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article. #
Answer
Radius = 3.5 cm, height = 10 cm.
TSA = Curved SA of cylinder + 2 × Curved SA of hemisphere
$$= 2\pi rh + 2 \times 2\pi r^2 = 2\pi r(h + 2r) = 2 \times \frac{22}{7} \times 3.5 \times (10 + 7) = 22 \times 17 = 374 \text{ cm}^2$$